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1.随机产生一个10×10的矩阵，矩阵元素为0～10之间的数；对每列求均值产生一个均值行向量；求矩阵的秩，判断是否为可逆阵，在可逆的情况下求解逆矩阵，并验证逆矩阵的正确性，即判断两个矩阵的乘积是否为单位阵。
clear all; close all; A = rand(10,10)*10 B = inv(A) disp(‘The A*B equals to :‘); disp(A*B); fprintf(‘sure det(A * B) == %f‘,det(A*B)); C = det(A*B)
clear all;
close all;
a = [1 -6 9];
b = [0 1 5];
c = [0 1 -4];
d = conv(conv(a,b),c)
disp(‘The roots is :‘);
AN = roots(d);
len = length(AN);
for i = 1:len
fprintf(‘X%d = %3f\n‘,i,AN(i));
end
3. 将有理分式进行部分分式展开:\(H(s) = \displaystyle\frac{2s^2+2s+13}{(s+1)(s-2)(s-3)}\)
clear all; close all; n = [2 2 13]; p1 = [0 1 1]; p2 = [0 1 -2]; p3 = [0 1 -3]; r = [-1 2 3]; p = poly(r) [above,below,mul] = residue(n,p)
4. 利用符号函数绘制信号\(f_1(t)=\displaystyle\frac{sin(\omega_0(t-t_0))}{\omega_0t}\), \(f_2(t) = \displaystyle\frac{sin\omega_0(t-t_0)}{\omega_0(t-t_0)}\),\(f(t) = f_1(t)+f_2(t)\)
OmigaSet = [2:3:15];
t = sym(‘t‘);
Omiga = sym(‘Omiga‘);
f1 = sym(‘sin(Omiga*t)/(Omiga*t)‘);
f2 = sym(‘sin(Omiga*(t-3))/(Omiga*(t-3))‘);
for i = 1:length(OmigaSet)
Plotf1 = subs(f1,Omiga,OmigaSet(i));
Plotf2 = subs(f2,Omiga,OmigaSet(i));
subplot(2,2,1),h1 = ezplot(Plotf1);
xlim([-3,7]);
ylim([-.5,1.5]);
subplot(2,2,2),h2 = ezplot(Plotf2);
xlim([-3,7]);
ylim([-.5,1.5]);
subplot(2,2,[3 4]),h3 = ezplot(Plotf1 + Plotf2);
xlim([-3,7]);
ylim([-.5,1.5]);
set([h1,h2,h3],‘LineWidth‘,1);
title(‘Superposition Of two figure above‘);
scrsz = get(0,‘ScreenSize‘);
set(gcf,‘Position‘,scrsz);
saveas(gcf,[‘Item2_4_‘ num2str(i) ‘.png‘]);
end
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原文地址：http://www.cnblogs.com/lccurious/p/5164424.html
